{"id":5278,"date":"2018-05-17T02:08:18","date_gmt":"2018-05-17T02:08:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/coreq-collegealgebra\/?post_type=chapter&#038;p=5278"},"modified":"2018-05-21T17:58:00","modified_gmt":"2018-05-21T17:58:00","slug":"exponential-growth-and-decay","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/exponential-growth-and-decay\/","title":{"raw":"Exponential Growth and Decay*","rendered":"Exponential Growth and Decay*"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Graph exponential growth and decay<\/li>\r\n \t<li>Solve problems involving radioactive decay, carbon dating, and half life<\/li>\r\n \t<li>Use Newton's law of cooling<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:\r\n<p style=\"text-align: center;\">[latex]y={A}_{0}{e}^{kt}[\/latex]<\/p>\r\nwhere [latex]{A}_{0}[\/latex] is equal to the value at time zero, <em>e<\/em>\u00a0is Euler\u2019s constant, and <em>k<\/em>\u00a0is a positive constant that determines the rate (percentage) of growth. We may use the <strong>exponential growth<\/strong> function in applications involving <strong>doubling time<\/strong>, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.\r\n\r\nOn the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the <strong>exponential decay<\/strong> model. Again, we have the form [latex]y={A}_{0}{e}^{-kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and <em>e<\/em>\u00a0is Euler\u2019s constant. Now <em>k<\/em>\u00a0is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating <strong>half-life<\/strong>, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.\r\n\r\nIn our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in the graphs below. It is important to remember that, although parts of each of the two graphs seem to lie on the <em>x<\/em>-axis, they are really a tiny distance above the <em>x<\/em>-axis.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181337\/CNX_Precalc_Figure_04_07_0022.jpg\" alt=\"Graph of y=2e^(3x) with the labeled points (-1\/3, 2\/e), (0, 2), and (1\/3, 2e) and with the asymptote at y=0.\" width=\"487\" height=\"326\" data-media-type=\"image\/jpg\" \/> A graph showing exponential growth. The equation is [latex]y=2{e}^{3x}[\/latex].[\/caption][caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181339\/CNX_Precalc_Figure_04_07_0032.jpg\" alt=\"Graph of y=3e^(-2x) with the labeled points (-1\/2, 3e), (0, 3), and (1\/2, 3\/e) and with the asymptote at y=0.\" width=\"487\" height=\"438\" data-media-type=\"image\/jpg\" \/> A graph showing exponential decay. The equation is [latex]y=3{e}^{-2x}[\/latex].[\/caption]&nbsp;\r\n\r\nExponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The <strong>order of magnitude<\/strong> is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, <strong>Proxima Centauri<\/strong>, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is [latex]4.01134972\\times {10}^{13}[\/latex]. So, we could describe this number as having order of magnitude [latex]{10}^{13}[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Characteristics of the Exponential Function, [latex]y=A_{0}e^{kt}[\/latex]<\/h3>\r\nAn exponential function with the form [latex]y={A}_{0}{e}^{kt}[\/latex] has the following characteristics:\r\n<ul>\r\n \t<li>one-to-one function<\/li>\r\n \t<li>horizontal asymptote: <em>y\u00a0<\/em>= 0<\/li>\r\n \t<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\r\n \t<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\r\n \t<li>x intercept: none<\/li>\r\n \t<li>y-intercept: [latex]\\left(0,{A}_{0}\\right)[\/latex]<\/li>\r\n \t<li>increasing if <em>k\u00a0<\/em>&gt; 0<\/li>\r\n \t<li>decreasing if <em>k\u00a0<\/em>&lt; 0<\/li>\r\n<\/ul>\r\nAn exponential function models exponential growth when k\u00a0&gt; 0 and exponential decay when k\u00a0&lt; 0.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe still graphs below include important characteristics of an exponential function that can\u00a0illustrate whether the function represents growth or decay, and the speed at which the function grows or decays.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181341\/CNX_Precalc_Figure_04_07_004new2.jpg\" alt=\"Two graphs of y=(A_0)(e^(kt)) with the asymptote at y=0. The first graph is of when k&gt;0 and with the labeled points (1\/k, (A_0)e), (0, A_0), and (-1\/k, (A_0)\/e). The second graph is of when k&lt;0 and with the labeled points (-1\/k, (A_0)e), (0, A_0), and (1\/k, (A_0)\/e).\" width=\"731\" height=\"337\" data-media-type=\"image\/jpg\" \/>\r\n\r\nYour task is to add the key features in the still graphs above to the Desmos graph below by using sliders where applicable. The function [latex]f(x) = a\\cdot{e}^{kx}[\/latex] has been included.\r\n\r\nInclude the following key features:\r\n<ul>\r\n \t<li>The\u00a0horizontal asymptote<\/li>\r\n \t<li>y-intercept<\/li>\r\n \t<li>The key points labeled in the still graphs above<\/li>\r\n<\/ul>\r\nhttps:\/\/www.desmos.com\/calculator\/4zae7vnjfr\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing Exponential Growth<\/h3>\r\nA population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.\r\n\r\n[reveal-answer q=\"151444\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"151444\"]\r\n\r\nWhen an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[\/latex] we use the fact that [latex]{A}_{0}[\/latex] is the amount at time zero, so [latex]{A}_{0}=10[\/latex]. To find <em>k<\/em>, use the fact that after one hour [latex]\\left(t=1\\right)[\/latex] the population doubles from 10 to 20. The formula is derived as follows\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }20=10{e}^{k\\cdot 1}\\hfill &amp; \\hfill \\\\ \\text{ }2={e}^{k}\\hfill &amp; \\text{Divide by 10}\\hfill \\\\ \\mathrm{ln}2=k\\hfill &amp; \\text{Take the natural logarithm}\\hfill \\end{array}[\/latex]<\/p>\r\nso [latex]k=\\mathrm{ln}\\left(2\\right)[\/latex]. Thus the equation we want to graph is [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}=10{\\left({e}^{\\mathrm{ln}2}\\right)}^{t}=10\\cdot {2}^{t}[\/latex]. The graph is shown below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181344\/CNX_Precalc_Figure_04_07_0052.jpg\" alt=\"A graph starting at ten on the y-axis and rising rapidly to the right.\" width=\"487\" height=\"438\" data-media-type=\"image\/jpg\" \/> The graph of [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}[\/latex]<strong>\u00a0<\/strong>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nThe population of bacteria after ten hours is 10,240. We could describe this amount is being of the order of magnitude [latex]{10}^{4}[\/latex]. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude [latex]{10}^{7}[\/latex], so we could say that the population has increased by three orders of magnitude in ten hours.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Calculating Doubling Time<\/h2>\r\nFor growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the <strong>doubling time<\/strong>.\r\n\r\nGiven the basic <strong>exponential growth<\/strong> equation [latex]A={A}_{0}{e}^{kt}[\/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[\/latex].\r\n\r\nThe formula is derived as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2{A}_{0}={A}_{0}{e}^{kt}\\hfill &amp; \\hfill \\\\ 2={e}^{kt}\\hfill &amp; \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}2=kt\\hfill &amp; \\text{Take the natural logarithm}.\\hfill \\\\ t=\\frac{\\mathrm{ln}2}{k}\\hfill &amp; \\text{Divide by the coefficient of }t.\\hfill \\end{array}[\/latex]<\/p>\r\nThus the doubling time is\r\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}2}{k}[\/latex]<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Function That Describes Exponential Growth<\/h3>\r\nAccording to Moore\u2019s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.\r\n\r\n[reveal-answer q=\"567900\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"567900\"]\r\n\r\nThe formula is derived as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{\\mathrm{ln}2}{k}\\hfill &amp; \\text{The doubling time formula}.\\hfill \\\\ 2=\\frac{\\mathrm{ln}2}{k}\\hfill &amp; \\text{Use a doubling time of two years}.\\hfill \\\\ k=\\frac{\\mathrm{ln}2}{2}\\hfill &amp; \\text{Multiply by }k\\text{ and divide by 2}.\\hfill \\\\ A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}\\hfill &amp; \\text{Substitute }k\\text{ into the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe function is [latex]A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nRecent data suggests that, as of 2013, the rate of growth predicted by Moore\u2019s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.\r\n\r\n[reveal-answer q=\"271465\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"271465\"][latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129930&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Half-Life<\/h2>\r\nWe now turn to <strong>exponential decay<\/strong>. One of the common terms associated with exponential decay, as stated above, is <strong>half-life<\/strong>, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.\r\n\r\nTo find the half-life of a function describing exponential decay, solve the following equation:\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[\/latex]<\/p>\r\nWe find that the half-life depends only on the constant <em>k<\/em>\u00a0and not on the starting quantity [latex]{A}_{0}[\/latex].\r\n\r\nThe formula is derived as follows\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}\\hfill &amp; \\hfill \\\\ \\frac{1}{2}={e}^{kt}\\hfill &amp; \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{1}{2}\\right)=kt\\hfill &amp; \\text{Take the natural log}.\\hfill \\\\ -\\mathrm{ln}\\left(2\\right)=kt\\hfill &amp; \\text{Apply laws of logarithms}.\\hfill \\\\ -\\frac{\\mathrm{ln}\\left(2\\right)}{k}=t\\hfill &amp; \\text{Divide by }k.\\hfill \\end{array}[\/latex]<\/p>\r\nSince <em>t<\/em>, the time, is positive, <em>k<\/em>\u00a0must, as expected, be negative. This gives us the half-life formula\r\n<p style=\"text-align: center;\">[latex]t=-\\frac{\\mathrm{ln}\\left(2\\right)}{k}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">Radioactive decay<\/p>\r\nIn previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.\r\n\r\nOne such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.\r\n<table summary=\"Seven rows and three columns. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center;\" data-align=\"center\">Substance<\/th>\r\n<th style=\"text-align: center;\" data-align=\"center\">Use<\/th>\r\n<th style=\"text-align: center;\" data-align=\"center\">Half-life<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>gallium-67<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>80 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>cobalt-60<\/td>\r\n<td>manufacturing<\/td>\r\n<td>5.3 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>technetium-99m<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>6 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>americium-241<\/td>\r\n<td>construction<\/td>\r\n<td>432 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>carbon-14<\/td>\r\n<td>archeological dating<\/td>\r\n<td>5,715 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>uranium-235<\/td>\r\n<td>atomic power<\/td>\r\n<td>703,800,000 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t}\\hfill \\\\ A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}}\\hfill \\end{array}[\/latex]<\/p>\r\nwhere\r\n<ul>\r\n \t<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\r\n \t<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\r\n \t<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\r\n \t<li><em>y<\/em>\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\r\n<\/ul>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>How To: Given the half-life, find the decay rate.<\/h3>\r\n<ol>\r\n \t<li>Write [latex]A={A}_{o}{e}^{kt}[\/latex].<\/li>\r\n \t<li>Replace <em>A<\/em>\u00a0by [latex]\\frac{1}{2}{A}_{0}[\/latex] and replace <em>t<\/em>\u00a0by the given half-life.<\/li>\r\n \t<li>Solve to find <em>k<\/em>. Express <em>k<\/em>\u00a0as an exact value (do not round).<\/li>\r\n<\/ol>\r\nNote: <em>It is also possible to find the decay rate using<\/em> [latex]k=-\\frac{\\mathrm{ln}\\left(2\\right)}{t}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Formula for Radioactive Decay to Find the Quantity of a Substance<\/h3>\r\nHow long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?\r\n\r\n[reveal-answer q=\"536683\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"536683\"]\r\n\r\n[latex]\\begin{array}{l}\\text{ }y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill &amp; \\hfill \\\\ \\text{ }900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill &amp; \\text{After 10% decays, 900 grams are left}.\\hfill \\\\ \\text{ }0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill &amp; \\text{Divide by 1000}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)\\hfill &amp; \\text{Take ln of both sides}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill &amp; \\text{ln}\\left({e}^{M}\\right)=M\\hfill \\\\ \\text{ }\\text{ }t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}\\begin{array}{l}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Solve for }t.\\hfill \\\\ \\text{ }\\text{ }t\\approx \\text{106,979,777 years}\\hfill &amp; \\hfill \\end{array}[\/latex]\r\n<h4>Analysis of the Solution<\/h4>\r\nTen percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nHow long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?\r\n\r\n[reveal-answer q=\"923702\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"923702\"][latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Function that Describes Radioactive Decay<\/h3>\r\nThe half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, <em>t<\/em>.\r\n\r\n[reveal-answer q=\"594419\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"594419\"]\r\n\r\nThis formula is derived as follows.\r\n\r\n[latex]\\begin{array}{l}\\text{ }A={A}_{0}{e}^{kt}\\hfill &amp; \\text{The continuous growth formula}.\\hfill \\\\ 0.5{A}_{0}={A}_{0}{e}^{k\\cdot 5730}\\hfill &amp; \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }f\\left(t\\right).\\hfill \\\\ \\text{ }0.5={e}^{5730k}\\hfill &amp; \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)=5730k\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\hfill &amp; \\text{Divide by the coefficient of }k.\\hfill \\\\ \\text{ }A={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}\\hfill &amp; \\text{Substitute for }r\\text{ in the continuous growth formula}.\\hfill \\end{array}[\/latex]\r\n\r\nThe function that describes this continuous decay is [latex]f\\left(t\\right)={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]. We observe that the coefficient of <em>t<\/em>, [latex]\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\approx -1.2097[\/latex] is negative, as expected in the case of exponential decay.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of carbon-14 remaining as a function of time, measured in years.\r\n\r\n[reveal-answer q=\"267261\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"267261\"]\r\n\r\n[latex]f\\left(t\\right)={A}_{0}{e}^{-0.0000000087t}[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom150\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=100026&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Radiocarbon Dating<\/h2>\r\nThe formula for radioactive decay is important in <strong>radiocarbon dating<\/strong>, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.\r\n\r\nCarbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates\u2014although the ratio has changed slightly over the centuries.\r\n\r\nAs long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.\r\n\r\nSince the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after <em>t<\/em>\u00a0years is\r\n<p style=\"text-align: center;\">[latex]A\\approx {A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]<\/p>\r\nwhere\r\n<ul>\r\n \t<li><em>A<\/em>\u00a0is the amount of carbon-14 remaining<\/li>\r\n \t<li>[latex]{A}_{0}[\/latex] is the amount of carbon-14 when the plant or animal began decaying.<\/li>\r\n<\/ul>\r\nThis formula is derived as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }A={A}_{0}{e}^{kt}\\hfill &amp; \\text{The continuous growth formula}.\\hfill \\\\ \\text{ }0.5{A}_{0}={A}_{0}{e}^{k\\cdot 5730}\\hfill &amp; \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }f\\left(t\\right).\\hfill \\\\ \\text{ }0.5={e}^{5730k}\\hfill &amp; \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)=5730k\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\hfill &amp; \\text{Divide by the coefficient of }k.\\hfill \\\\ \\text{ }A={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}\\hfill &amp; \\text{Substitute for }r\\text{ in the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\r\nTo find the age of an object, we solve this equation for <em>t<\/em>:\r\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(\\frac{A}{{A}_{0}}\\right)}{-0.000121}[\/latex]<\/p>\r\nOut of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let <em>r<\/em>\u00a0be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation [latex]A\\approx {A}_{0}{e}^{-0.000121t}[\/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\\frac{A}{{A}_{0}}\\approx {e}^{-0.000121t}[\/latex]. We solve this equation for <em>t<\/em>, to get\r\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>How To: Given the percentage of carbon-14 in an object, determine its age.<\/h3>\r\n<ol>\r\n \t<li>Express the given percentage of carbon-14 as an equivalent decimal, <em>k<\/em>.<\/li>\r\n \t<li>Substitute for <em>k<\/em> in the equation [latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex] and solve for the age, <em>t<\/em>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Age of a Bone<\/h3>\r\nA bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?\r\n\r\n[reveal-answer q=\"954630\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"954630\"]\r\n\r\nWe substitute 20% = 0.20 for <em>k<\/em>\u00a0in the equation and solve for <em>t<\/em>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}\\hfill &amp; \\text{Use the general form of the equation}.\\hfill \\\\ =\\frac{\\mathrm{ln}\\left(0.20\\right)}{-0.000121}\\hfill &amp; \\text{Substitute for }r.\\hfill \\\\ \\approx 13301\\hfill &amp; \\text{Round to the nearest year}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe bone fragment is about 13,301 years old.\r\n<h4>Analysis of the Solution<\/h4>\r\nThe instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as [latex]\\text{13,301 years}\\pm \\text{1% or 13,301 years}\\pm \\text{133 years}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nCesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?\r\n\r\n[reveal-answer q=\"758746\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"758746\"]less than 230 years, 229.3157 to be exact[\/hidden-answer]\r\n<iframe id=\"mom15\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=29686&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Graph exponential growth and decay<\/li>\n<li>Solve problems involving radioactive decay, carbon dating, and half life<\/li>\n<li>Use Newton&#8217;s law of cooling<\/li>\n<\/ul>\n<\/div>\n<p>In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:<\/p>\n<p style=\"text-align: center;\">[latex]y={A}_{0}{e}^{kt}[\/latex]<\/p>\n<p>where [latex]{A}_{0}[\/latex] is equal to the value at time zero, <em>e<\/em>\u00a0is Euler\u2019s constant, and <em>k<\/em>\u00a0is a positive constant that determines the rate (percentage) of growth. We may use the <strong>exponential growth<\/strong> function in applications involving <strong>doubling time<\/strong>, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.<\/p>\n<p>On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the <strong>exponential decay<\/strong> model. Again, we have the form [latex]y={A}_{0}{e}^{-kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and <em>e<\/em>\u00a0is Euler\u2019s constant. Now <em>k<\/em>\u00a0is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating <strong>half-life<\/strong>, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.<\/p>\n<p>In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in the graphs below. It is important to remember that, although parts of each of the two graphs seem to lie on the <em>x<\/em>-axis, they are really a tiny distance above the <em>x<\/em>-axis.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181337\/CNX_Precalc_Figure_04_07_0022.jpg\" alt=\"Graph of y=2e^(3x) with the labeled points (-1\/3, 2\/e), (0, 2), and (1\/3, 2e) and with the asymptote at y=0.\" width=\"487\" height=\"326\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">A graph showing exponential growth. The equation is [latex]y=2{e}^{3x}[\/latex].<\/p>\n<\/div>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181339\/CNX_Precalc_Figure_04_07_0032.jpg\" alt=\"Graph of y=3e^(-2x) with the labeled points (-1\/2, 3e), (0, 3), and (1\/2, 3\/e) and with the asymptote at y=0.\" width=\"487\" height=\"438\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">A graph showing exponential decay. The equation is [latex]y=3{e}^{-2x}[\/latex].<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The <strong>order of magnitude<\/strong> is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, <strong>Proxima Centauri<\/strong>, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is [latex]4.01134972\\times {10}^{13}[\/latex]. So, we could describe this number as having order of magnitude [latex]{10}^{13}[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Characteristics of the Exponential Function, [latex]y=A_{0}e^{kt}[\/latex]<\/h3>\n<p>An exponential function with the form [latex]y={A}_{0}{e}^{kt}[\/latex] has the following characteristics:<\/p>\n<ul>\n<li>one-to-one function<\/li>\n<li>horizontal asymptote: <em>y\u00a0<\/em>= 0<\/li>\n<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\n<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\n<li>x intercept: none<\/li>\n<li>y-intercept: [latex]\\left(0,{A}_{0}\\right)[\/latex]<\/li>\n<li>increasing if <em>k\u00a0<\/em>&gt; 0<\/li>\n<li>decreasing if <em>k\u00a0<\/em>&lt; 0<\/li>\n<\/ul>\n<p>An exponential function models exponential growth when k\u00a0&gt; 0 and exponential decay when k\u00a0&lt; 0.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The still graphs below include important characteristics of an exponential function that can\u00a0illustrate whether the function represents growth or decay, and the speed at which the function grows or decays.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181341\/CNX_Precalc_Figure_04_07_004new2.jpg\" alt=\"Two graphs of y=(A_0)(e^(kt)) with the asymptote at y=0. The first graph is of when k&gt;0 and with the labeled points (1\/k, (A_0)e), (0, A_0), and (-1\/k, (A_0)\/e). The second graph is of when k&lt;0 and with the labeled points (-1\/k, (A_0)e), (0, A_0), and (1\/k, (A_0)\/e).\" width=\"731\" height=\"337\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>Your task is to add the key features in the still graphs above to the Desmos graph below by using sliders where applicable. The function [latex]f(x) = a\\cdot{e}^{kx}[\/latex] has been included.<\/p>\n<p>Include the following key features:<\/p>\n<ul>\n<li>The\u00a0horizontal asymptote<\/li>\n<li>y-intercept<\/li>\n<li>The key points labeled in the still graphs above<\/li>\n<\/ul>\n<p>https:\/\/www.desmos.com\/calculator\/4zae7vnjfr<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing Exponential Growth<\/h3>\n<p>A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q151444\">Solution<\/span><\/p>\n<div id=\"q151444\" class=\"hidden-answer\" style=\"display: none\">\n<p>When an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[\/latex] we use the fact that [latex]{A}_{0}[\/latex] is the amount at time zero, so [latex]{A}_{0}=10[\/latex]. To find <em>k<\/em>, use the fact that after one hour [latex]\\left(t=1\\right)[\/latex] the population doubles from 10 to 20. The formula is derived as follows<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }20=10{e}^{k\\cdot 1}\\hfill & \\hfill \\\\ \\text{ }2={e}^{k}\\hfill & \\text{Divide by 10}\\hfill \\\\ \\mathrm{ln}2=k\\hfill & \\text{Take the natural logarithm}\\hfill \\end{array}[\/latex]<\/p>\n<p>so [latex]k=\\mathrm{ln}\\left(2\\right)[\/latex]. Thus the equation we want to graph is [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}=10{\\left({e}^{\\mathrm{ln}2}\\right)}^{t}=10\\cdot {2}^{t}[\/latex]. The graph is shown below.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181344\/CNX_Precalc_Figure_04_07_0052.jpg\" alt=\"A graph starting at ten on the y-axis and rising rapidly to the right.\" width=\"487\" height=\"438\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">The graph of [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}[\/latex]<strong>\u00a0<\/strong><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>The population of bacteria after ten hours is 10,240. We could describe this amount is being of the order of magnitude [latex]{10}^{4}[\/latex]. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude [latex]{10}^{7}[\/latex], so we could say that the population has increased by three orders of magnitude in ten hours.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Calculating Doubling Time<\/h2>\n<p>For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the <strong>doubling time<\/strong>.<\/p>\n<p>Given the basic <strong>exponential growth<\/strong> equation [latex]A={A}_{0}{e}^{kt}[\/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[\/latex].<\/p>\n<p>The formula is derived as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2{A}_{0}={A}_{0}{e}^{kt}\\hfill & \\hfill \\\\ 2={e}^{kt}\\hfill & \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}2=kt\\hfill & \\text{Take the natural logarithm}.\\hfill \\\\ t=\\frac{\\mathrm{ln}2}{k}\\hfill & \\text{Divide by the coefficient of }t.\\hfill \\end{array}[\/latex]<\/p>\n<p>Thus the doubling time is<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}2}{k}[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Function That Describes Exponential Growth<\/h3>\n<p>According to Moore\u2019s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567900\">Solution<\/span><\/p>\n<div id=\"q567900\" class=\"hidden-answer\" style=\"display: none\">\n<p>The formula is derived as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{\\mathrm{ln}2}{k}\\hfill & \\text{The doubling time formula}.\\hfill \\\\ 2=\\frac{\\mathrm{ln}2}{k}\\hfill & \\text{Use a doubling time of two years}.\\hfill \\\\ k=\\frac{\\mathrm{ln}2}{2}\\hfill & \\text{Multiply by }k\\text{ and divide by 2}.\\hfill \\\\ A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}\\hfill & \\text{Substitute }k\\text{ into the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\n<p>The function is [latex]A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Recent data suggests that, as of 2013, the rate of growth predicted by Moore\u2019s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q271465\">Solution<\/span><\/p>\n<div id=\"q271465\" class=\"hidden-answer\" style=\"display: none\">[latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129930&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Half-Life<\/h2>\n<p>We now turn to <strong>exponential decay<\/strong>. One of the common terms associated with exponential decay, as stated above, is <strong>half-life<\/strong>, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.<\/p>\n<p>To find the half-life of a function describing exponential decay, solve the following equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[\/latex]<\/p>\n<p>We find that the half-life depends only on the constant <em>k<\/em>\u00a0and not on the starting quantity [latex]{A}_{0}[\/latex].<\/p>\n<p>The formula is derived as follows<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}\\hfill & \\hfill \\\\ \\frac{1}{2}={e}^{kt}\\hfill & \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{1}{2}\\right)=kt\\hfill & \\text{Take the natural log}.\\hfill \\\\ -\\mathrm{ln}\\left(2\\right)=kt\\hfill & \\text{Apply laws of logarithms}.\\hfill \\\\ -\\frac{\\mathrm{ln}\\left(2\\right)}{k}=t\\hfill & \\text{Divide by }k.\\hfill \\end{array}[\/latex]<\/p>\n<p>Since <em>t<\/em>, the time, is positive, <em>k<\/em>\u00a0must, as expected, be negative. This gives us the half-life formula<\/p>\n<p style=\"text-align: center;\">[latex]t=-\\frac{\\mathrm{ln}\\left(2\\right)}{k}[\/latex]<\/p>\n<p style=\"text-align: center;\">Radioactive decay<\/p>\n<p>In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/p>\n<p>One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\n<table summary=\"Seven rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th style=\"text-align: center;\" data-align=\"center\">Substance<\/th>\n<th style=\"text-align: center;\" data-align=\"center\">Use<\/th>\n<th style=\"text-align: center;\" data-align=\"center\">Half-life<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>gallium-67<\/td>\n<td>nuclear medicine<\/td>\n<td>80 hours<\/td>\n<\/tr>\n<tr>\n<td>cobalt-60<\/td>\n<td>manufacturing<\/td>\n<td>5.3 years<\/td>\n<\/tr>\n<tr>\n<td>technetium-99m<\/td>\n<td>nuclear medicine<\/td>\n<td>6 hours<\/td>\n<\/tr>\n<tr>\n<td>americium-241<\/td>\n<td>construction<\/td>\n<td>432 years<\/td>\n<\/tr>\n<tr>\n<td>carbon-14<\/td>\n<td>archeological dating<\/td>\n<td>5,715 years<\/td>\n<\/tr>\n<tr>\n<td>uranium-235<\/td>\n<td>atomic power<\/td>\n<td>703,800,000 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t}\\hfill \\\\ A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}}\\hfill \\end{array}[\/latex]<\/p>\n<p>where<\/p>\n<ul>\n<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\n<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\n<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\n<li><em>y<\/em>\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the half-life, find the decay rate.<\/h3>\n<ol>\n<li>Write [latex]A={A}_{o}{e}^{kt}[\/latex].<\/li>\n<li>Replace <em>A<\/em>\u00a0by [latex]\\frac{1}{2}{A}_{0}[\/latex] and replace <em>t<\/em>\u00a0by the given half-life.<\/li>\n<li>Solve to find <em>k<\/em>. Express <em>k<\/em>\u00a0as an exact value (do not round).<\/li>\n<\/ol>\n<p>Note: <em>It is also possible to find the decay rate using<\/em> [latex]k=-\\frac{\\mathrm{ln}\\left(2\\right)}{t}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Formula for Radioactive Decay to Find the Quantity of a Substance<\/h3>\n<p>How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q536683\">Solution<\/span><\/p>\n<div id=\"q536683\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\text{ }y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill & \\hfill \\\\ \\text{ }900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill & \\text{After 10% decays, 900 grams are left}.\\hfill \\\\ \\text{ }0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill & \\text{Divide by 1000}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill & \\text{ln}\\left({e}^{M}\\right)=M\\hfill \\\\ \\text{ }\\text{ }t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}\\begin{array}{l}{cccc}& & & \\end{array}\\hfill & \\text{Solve for }t.\\hfill \\\\ \\text{ }\\text{ }t\\approx \\text{106,979,777 years}\\hfill & \\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q923702\">Solution<\/span><\/p>\n<div id=\"q923702\" class=\"hidden-answer\" style=\"display: none\">[latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Function that Describes Radioactive Decay<\/h3>\n<p>The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, <em>t<\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q594419\">Solution<\/span><\/p>\n<div id=\"q594419\" class=\"hidden-answer\" style=\"display: none\">\n<p>This formula is derived as follows.<\/p>\n<p>[latex]\\begin{array}{l}\\text{ }A={A}_{0}{e}^{kt}\\hfill & \\text{The continuous growth formula}.\\hfill \\\\ 0.5{A}_{0}={A}_{0}{e}^{k\\cdot 5730}\\hfill & \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }f\\left(t\\right).\\hfill \\\\ \\text{ }0.5={e}^{5730k}\\hfill & \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)=5730k\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\hfill & \\text{Divide by the coefficient of }k.\\hfill \\\\ \\text{ }A={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}\\hfill & \\text{Substitute for }r\\text{ in the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\n<p>The function that describes this continuous decay is [latex]f\\left(t\\right)={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]. We observe that the coefficient of <em>t<\/em>, [latex]\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\approx -1.2097[\/latex] is negative, as expected in the case of exponential decay.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of carbon-14 remaining as a function of time, measured in years.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q267261\">Solution<\/span><\/p>\n<div id=\"q267261\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(t\\right)={A}_{0}{e}^{-0.0000000087t}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom150\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=100026&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Radiocarbon Dating<\/h2>\n<p>The formula for radioactive decay is important in <strong>radiocarbon dating<\/strong>, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.<\/p>\n<p>Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates\u2014although the ratio has changed slightly over the centuries.<\/p>\n<p>As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.<\/p>\n<p>Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after <em>t<\/em>\u00a0years is<\/p>\n<p style=\"text-align: center;\">[latex]A\\approx {A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]<\/p>\n<p>where<\/p>\n<ul>\n<li><em>A<\/em>\u00a0is the amount of carbon-14 remaining<\/li>\n<li>[latex]{A}_{0}[\/latex] is the amount of carbon-14 when the plant or animal began decaying.<\/li>\n<\/ul>\n<p>This formula is derived as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }A={A}_{0}{e}^{kt}\\hfill & \\text{The continuous growth formula}.\\hfill \\\\ \\text{ }0.5{A}_{0}={A}_{0}{e}^{k\\cdot 5730}\\hfill & \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }f\\left(t\\right).\\hfill \\\\ \\text{ }0.5={e}^{5730k}\\hfill & \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)=5730k\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\hfill & \\text{Divide by the coefficient of }k.\\hfill \\\\ \\text{ }A={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}\\hfill & \\text{Substitute for }r\\text{ in the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\n<p>To find the age of an object, we solve this equation for <em>t<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(\\frac{A}{{A}_{0}}\\right)}{-0.000121}[\/latex]<\/p>\n<p>Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let <em>r<\/em>\u00a0be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation [latex]A\\approx {A}_{0}{e}^{-0.000121t}[\/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\\frac{A}{{A}_{0}}\\approx {e}^{-0.000121t}[\/latex]. We solve this equation for <em>t<\/em>, to get<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the percentage of carbon-14 in an object, determine its age.<\/h3>\n<ol>\n<li>Express the given percentage of carbon-14 as an equivalent decimal, <em>k<\/em>.<\/li>\n<li>Substitute for <em>k<\/em> in the equation [latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex] and solve for the age, <em>t<\/em>.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Age of a Bone<\/h3>\n<p>A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q954630\">Solution<\/span><\/p>\n<div id=\"q954630\" class=\"hidden-answer\" style=\"display: none\">\n<p>We substitute 20% = 0.20 for <em>k<\/em>\u00a0in the equation and solve for <em>t<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}\\hfill & \\text{Use the general form of the equation}.\\hfill \\\\ =\\frac{\\mathrm{ln}\\left(0.20\\right)}{-0.000121}\\hfill & \\text{Substitute for }r.\\hfill \\\\ \\approx 13301\\hfill & \\text{Round to the nearest year}.\\hfill \\end{array}[\/latex]<\/p>\n<p>The bone fragment is about 13,301 years old.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as [latex]\\text{13,301 years}\\pm \\text{1% or 13,301 years}\\pm \\text{133 years}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q758746\">Solution<\/span><\/p>\n<div id=\"q758746\" class=\"hidden-answer\" style=\"display: none\">less than 230 years, 229.3157 to be exact<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom15\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=29686&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5278\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Exponential Growth and Decay Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/4zae7vnjfr\">https:\/\/www.desmos.com\/calculator\/4zae7vnjfr<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Abramson, Jay et al. . <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at 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GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":160,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Abramson, Jay et al. \",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for 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